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Let $f$ be a continuous function on $[-1,1]$. Suppose $\exists C$, constant such that $0<C<1$, and $\int_{-1}^1 \frac{|f(t)|}{|t|^n} dt \leq C^n$, $\forall n$. Then I want to show $f \equiv 0$.

Naively I guess from $\lim_{n \rightarrow \infty} C^n=0$, but how one can prove $f\equiv 0$?

Trial vai comment from @Martin R

Assume $|f(x)| > \epsilon>0$ on some interval $[a,b]$, then \begin{align} \int_{-1}^1 \frac{|f(t)|}{|t|^n}dt = 2 \int_{0}^1 \frac{f(t)}{t^n} dt \geq 2 \epsilon \int_{0}^1 t^{-n} dt = \frac{2\epsilon}{n-1} \end{align} Hence \begin{align} \frac{2\epsilon}{n-1} \leq \int_{-1}^1 \frac{|f(t)|}{|t|^n}dt \leq C^n \end{align} For all $n$. Taking $n\rightarrow \infty$ it becomes $0\leq \int_{-1}^1 \frac{|f(t)|}{|t|^n} dt \leq 0$. But How $|f(t)|=0$?

phy_math
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  • Taking $n \to \infty$ gives $\int_{-1}^1 \frac{\vert f(t) \vert}{\vert t\vert } dt =0$. Since $f$ is continuous you straight away get that $f \equiv 0$. – JackT Jun 07 '21 at 13:47
  • @JackT What??? I follow what you said if we assume $\infty=1$. – David C. Ullrich Jun 07 '21 at 14:01
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    Assume that $|f(x)| > \epsilon > 0$ on some interval $[a, b]$. Use that to compute a lower bound for $\int_{-1}^1 \frac{|f(t)|}{|t|^n} dt$. – Martin R Jun 07 '21 at 14:06
  • @MartinR, I update my trial on your comment. But still I do not know $|f(x)|=0$. – phy_math Jun 07 '21 at 15:44
  • @David C. Ullrich ah whoops I missed the that denominator was $\vert t\vert^n$ not $\vert t\vert$. Sorry about that. – JackT Jun 07 '21 at 20:43

2 Answers2

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Assume that $f$ is not identically zero. Then $|f(x)| \ge \epsilon > 0$ on some interval $[a, b]$ of positive length. Without loss of generality we can assume that $0 < a < b < 1$. Then $$ C^n \ge \int_{-1}^1 \frac{|f(t)|}{|t|^n} \, dt \ge \int_a^b \frac{\epsilon}{t^n} \, dt = \frac{\epsilon}{n-1} \left( \frac{1}{a^{n-1}}- \frac{1}{b^{n-1}}\right) = \frac{\epsilon}{(n-1)a^{n-1}}\left( 1 - \left( \frac ab \right)^{n-1}\right) $$ for all $n \ge 2$, or $$ (n-1) C (aC)^{n-1} \ge \epsilon \left( 1 - \left( \frac ab \right)^{n-1}\right) \, . $$ That is not possible, because the left-hand side converges to zero for $n \to \infty$, whereas the right-hand side converges to $\epsilon > 0$.

Martin R
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It is enough to consider the integral over $I=[0,1]$. If $f\neq0$, then, by continuity of $f$, we may assume with out loss of generality that there is $0<t_0<1$ such that $f(t_0)\neq0$; furthermore, we may assume that for some $\alpha>0$ small enough, $J=[t_0, a_0+\alpha)\subset [0,1]$, and $|f(t)|>\frac12|f(t_0)|>0$ for all $t\in J$.

Then

$$\frac12|f(t_0)|\frac{t_0^{1-n}-(t_0+\alpha)^{1-n}}{n-1}=\frac{1}{2(n-1)}|f(t_0)|t^{1-n}_0\Big(1-\big(\frac{t_0+\alpha}{t_0}\big)^{1-n}\Big)\leq \int^1_{0}\frac{|f(t)|}{t^n}dt\leq C^n$$

Hence

$$0<\frac{1}{2}|f(t_0)|t_0\leq (n-1)C^nt^n_0\frac{1}{\Big(1-\big(\frac{t_0+\alpha}{t_0}\big)^{1-n}\Big)}\xrightarrow{n\rightarrow\infty}0$$ contradiction!

Similar argument works $f$ over $[-1,0]$.

Mittens
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    I did not see what you posted until now. I was working on my answer on paper and typing. If the point are too important to you, I will remove my solution. The argument, I think, it pretty standard and anyone else would have done something very similar. – Mittens Jun 07 '21 at 17:00